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A 2.98 nF parallel-plate capacitor is charged to an initial potential difference of 49 V and then isolated. The dielectric material between the plates has a dielectric constant of 3.1. What is the potential difference of the capacitor after the dielectric is withdrawn? Answer in units of V.

Respuesta :

Answer:

Potential difference will be 151.9 volt  

Explanation:

We have given capacitance of the capacitor [tex]C=2.98nF=2.98\times 10^{-9}F[/tex]

Voltage V = 49 Volt

Dielectric constant K = 3.1

We have to find the potential difference

We know that when a dielectric medium is introduced then p[otential difference is increases by k times

As the dielectric constant k = 3.1

So potential difference will be = 3.1×49 = 151.9 volt

Answer:151.9 V

Explanation:

Given

Capacitance (C)=2.98 nF

Potential difference=49 V

dielectric strength k=3.1

Charge remains same after the removal of dielectric

thus[tex]Q_1=Q_2[/tex]

[tex]Q_1=2.98\times 49[/tex]

Now dielectric is removed so capacitance decreases to [tex]\frac{C}{k}[/tex]

[tex]Q_2=\frac{2.98}{3.1}\times V[/tex]

[tex]V=49\times 3.1=151.9 V[/tex]

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